Question

The ${{K}_{\beta }}$ X-ray of argon has a wavelength of 0.36nm. The minimum energy needed to ionize an argon atom is 16eV. Find the energy needed to knockout an electron from the K-shell of an argon atom.

Answer 1

Hint: From the given value of wavelength, we could find the energy of emitted X-ray by using the expression for energy in terms of wavelength. Also, make sure that you use the value of Planck’s constant according to the unit in which the energy has to be found. Now, you could find the sum of this energy and ionization energy to find the total knockout energy.
Formula used:
Expression for energy,
$\Rightarrow E=h\dfrac{c}{\lambda }$

Complete answer:
In the question, we are given the wavelength of the ${{K}_{\beta }}$ X-ray of argon as 0.36nm and 16eV is the minimum energy required by an argon atom to ionize. We are asked to find out the energy required to knockout an electron from the K-shell of an argon atom.
As a first step let us list out the given values.
Wavelength of ${{K}_{\beta }}$ X-ray is given by,
$\lambda =0.36nm=0.36\times {{10}^{-9}}m$ ………………………………… (1)
The ionization energy of an argon atom is given by,
${{E}_{i}}=16eV$ ……………………………. (2)
We know that, energy of each ${{K}_{\beta }}$ X-ray emitted is given by the expression,
$E=h\nu $
But we know,
$\nu =\dfrac{c}{\lambda }$
$\Rightarrow E=h\dfrac{c}{\lambda }$ ……………………………. (3)
Where, h is the Planck constant given by,
$h=4.14\times {{10}^{-15}}eVH{{z}^{-1}}$ ………………………….. (4)
$c$is the universal speed of electromagnetic radiation given by,
$c=3\times {{10}^{8}}m{{s}^{-1}}$ ……………………………… (5)
Substituting (1), (4) and (5) in (3), we get,
$E=\dfrac{4.14\times {{10}^{-15}}\times 3\times {{10}^{8}}}{0.36\times {{10}^{-9}}}$
$\Rightarrow E=34.5\times {{10}^{2}}eV$
$\Rightarrow E=3450eV$ ……………………………. (6)
The total energy required to knock out an electron from the K shell of an argon atom is the sum of both ionization energy of an argon atom and also the energy of the ${{K}_{\beta }}$ x-ray emitted.
So, in order to find the total knockout energy ${{E}_{k}}$ we have to find the sum of (2) and (6). That is,
${{E}_{k}}={{E}_{i}}+E$
$\Rightarrow {{E}_{k}}=16eV+3450eV$
$\Rightarrow {{E}_{k}}=3466eV$
Therefore, the energy needed to knockout an electron from the K-shell of an argon atom is $3.46KeV$

Note:
Make sure that you convert the given value of wavelength into meters. Since the last step involves summing the given ionization energy and the energy of an X-ray emitted, both these energies should be in the same units. So accordingly choose the value for Planck’s constant. Here, we have used the value in eV.